Computational Physics - Department of Physics

7.9 Exercises 235

IF( interpol > 1 )THEN

CALLpolint(hh,value(:,l),interpol,0.d0 ,inp,err2)

err1=MAX(err1,err2)

WRITE(6,'(D12.6,2X,D12.6,2X,D12.6)') inp, d(l), err1

ELSE

WRITE(6,'(D12.6,2X,D12.6,2X,D12.6)') d(l), d(l), err1

ENDIF

ENDDO

DEALLOCATE( w, r, d, e)

ENDDO

7.9 Exercises.

7.1.The aim of this problem is to solve Schrödinger’s equation for two electrons in a three-
dimensional harmonic oscillator well with and without a repulsive Coulomb interaction. Your
task is to solve this equation by reformulating it in a discretized form as an eigenvalue equa-
tion to be solved with Jacobi’s method. To achieve this you will have to write your own code
which implements Jacobi’s method.
Electrons confined in small areas in semiconductors, so-called quantum dots, form a hot
research area in modern solid-state physics, with applications spanning from such diverse
fields as quantum nano-medicine to the contruction of quantum gates.
Here we will assume that these electrons move in a three-dimensional harmonic oscillator
potential (they are confined by for example quadrupole fields) and repel each other via the
static Colulomb interaction. We assume spherical symmetry.
We are first interested in the solution of the radial part of Schrödinger’s equation for one
electron. This equation reads

− ̄

h^2

2 m

(

1

r^2

d

dr

r^2

d

dr

−

l(l+ 1 )

r^2

)

R(r)+V(r)R(r) =E R(r).

In our caseV(r)is the harmonic oscillator potential( 1 / 2 )kr^2 withk=mω^2 andEis the energy
of the harmonic oscillator in three dimensions. The oscillator frequency isωand the energies
are

Enl=h ̄ω

(

2 n+l+

3

2

)

,

withn= 0 , 1 , 2 ,...andl= 0 , 1 , 2 ,....
Since we have made a transformation to spherical coordinates it means thatr∈[ 0 ,∞).
The quantum numberlis the orbital momentum of the electron. Then we substituteR(r) =
( 1 /r)u(r)and obtain

− ̄

h^2

2 m

d^2

dr^2

u(r)+

(

V(r)+

l(l+ 1 )

r^2

h ̄^2

2 m

)

u(r) =E u(r).

The boundary conditions areu( 0 ) = 0 andu(∞) = 0.
We introduce a dimensionless variableρ= ( 1 /α)rwhereαis a constant with dimension
length and get

−

h ̄^2

2 mα^2

d^2

dρ^2

u(ρ)+

(

V(ρ)+

l(l+ 1 )

ρ^2

h ̄^2

2 mα^2

)

u(ρ) =E u(ρ).

We will set in this projectl= 0. InsertingV(ρ) = ( 1 / 2 )kα^2 ρ^2 we end up with